Let $g(x)=\dfrac{\sin(x)}{e^x}$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos(x)-\sin(x)}{e^x}$ (Choice B) B $\cos(x)-e^x$ (Choice C) C $\dfrac{\cos(x)}{e^x}$ (Choice D) D $e^x(\cos(x)-\sin(x))$
Answer: $g(x)$ is the quotient of two, more basic, expressions: $\sin(x)$ and $e^x$. Therefore, the derivative of $g$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = g ′ ( x ) = d d x ( sin ( x ) e x ) = d d x ( sin ( x ) ) e x − sin ( x ) d d x ( e x ) ( e x ) 2 = cos ( x ) ⋅ e x − sin ( x ) ⋅ e x ( e x ) 2 = e x ( cos ( x ) − sin ( x ) ) ( e x ) 2 = cos ( x ) − sin ( x ) e x The quotient rule Differentiate sin ( x ) and e x Simplify \begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{\sin(x)}{e^x}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\sin(x))e^x-\sin(x)\dfrac{d}{dx}(e^x)}{(e^x)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\cos(x)\cdot e^x-\sin(x)\cdot e^x}{(e^x)^2}&&\gray{\text{Differentiate }\sin(x)\text{ and }e^x} \\\\ &=\dfrac{\cancel{e^x}(\cos(x)-\sin(x))}{(e^x)^\cancel2}&&\gray{\text{Simplify}} \\\\ &=\dfrac{\cos(x)-\sin(x)}{e^x} \end{aligned} In conclusion, $g'(x)=\dfrac{\cos(x)-\sin(x)}{e^x}$